Chapter 6: Problem 22
Determine whether the set \(\mathcal{B}\) is a basis for the vector space \(V\) $$V=\mathscr{P}_{2}, \mathcal{B}=\left\\{x, 1-x, 1+x+x^{2}\right\\}$$
Short Answer
Expert verified
Yes, \( \mathcal{B} \) is a basis for \( V \).
Step by step solution
01
Define the Vector Space and Basis
The vector space \( V = \mathscr{P}_{2} \) consists of all polynomials with a degree of at most 2. A basis for a vector space is a set of vectors that are linearly independent and span the vector space. We need to determine if \( \mathcal{B} = \{x, 1-x, 1+x+x^{2}\} \) satisfies these conditions.
02
Check Linear Independence
To check if \( \mathcal{B} \) is linearly independent, we must determine if the equation \( a(x) + b(1-x) + c(1+x+x^{2}) = 0 \) has only the trivial solution \( a = b = c = 0 \). This can be expanded to \( (a-b+c)x + (b+c) + cx^{2} = 0 \). For the equation to hold for all \( x \), the coefficients of \( x^0 \), \( x^1 \), and \( x^2 \) must all be zero, leading to the system of equations: - \( a-b+c = 0 \) - \( b+c = 0 \) - \( c = 0 \).
03
Solve the System of Equations for Linear Independence
From the equation \( c = 0 \), we substitute \( c \) into the second equation to get \( b = 0 \). Substituting \( b = 0 \) and \( c = 0 \) into the first equation yields \( a = 0 \). Thus, \( a = b = c = 0 \) is the only solution, confirming \( \mathcal{B} \) is linearly independent.
04
Verify Spanning
Since \( \mathcal{B} \) is a set of 3 polynomials and \( \mathscr{P}_{2} \) is 3-dimensional, if \( \mathcal{B} \) is linearly independent, it must span the space. Therefore, the linear independence of \( \mathcal{B} \) also ensures it spans \( V \).
05
Conclusion
Since \( \mathcal{B} \) is both linearly independent and spans \( V = \mathscr{P}_{2} \), it is a basis for the vector space.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Space
A vector space is a collection of objects, which we call vectors, that can be added together and multiplied by scalars, which are numbers. To be a vector space, certain rules must be followed. For instance, the sum of any two vectors must also be a vector in the space, and scalar multiplication must behave predictably. In the case we're discussing, our vector space is \( \mathscr{P}_{2} \), which consists of all polynomials that have a degree of at most 2.
This means that any polynomial in this space can be written as \( ax^2 + bx + c \), where the coefficients \( a \), \( b \), and \( c \) are real numbers. These polynomials can represent vectors.
Understanding vector spaces provides a framework for studying polynomials using linear algebra, where operations are well-defined and structured.
This means that any polynomial in this space can be written as \( ax^2 + bx + c \), where the coefficients \( a \), \( b \), and \( c \) are real numbers. These polynomials can represent vectors.
Understanding vector spaces provides a framework for studying polynomials using linear algebra, where operations are well-defined and structured.
Linear Independence
Linear independence is a key concept in linear algebra. A set of vectors is considered linearly independent if no vector in the set is a linear combination of the others. This means you cannot write one vector as a sum of the multiples of the others in the set.
In the case of our set \( \mathcal{B} = \{x, 1-x, 1+x+x^{2}\} \), to check for linear independence, we see if the equation
If this system of coefficients leads to a trivial solution, where all coefficients are zero, the set is linearly independent.
In our example, solving the mathematical formulation resulted in \( a = b = c = 0 \), confirming that \( \mathcal{B} \) is indeed linearly independent.
In the case of our set \( \mathcal{B} = \{x, 1-x, 1+x+x^{2}\} \), to check for linear independence, we see if the equation
- \( a(x) + b(1-x) + c(1+x+x^{2}) = 0 \)
If this system of coefficients leads to a trivial solution, where all coefficients are zero, the set is linearly independent.
In our example, solving the mathematical formulation resulted in \( a = b = c = 0 \), confirming that \( \mathcal{B} \) is indeed linearly independent.
Basis
A basis of a vector space is a set of vectors that not only spans the space but is also linearly independent. This set allows us to express every vector in the space uniquely as a combination of basis vectors. For a vector space with polynomials of degree up to 2, such as \( \mathscr{P}_{2} \), a basis has exactly three vectors.
In simple terms:
In simple terms:
- Spanning means any vector in the space can be formed by a combination of vectors in the set.
- To confirm a set is a basis, check both linear independence and the ability to span the entire space.
Polynomials
Polynomials are algebraic expressions that involve sums of powers of a variable. They appear frequently in mathematics and are a fundamental part of linear algebra when dealing with polynomial vector spaces.
In our exercise, the vector space \( \mathscr{P}_{2} \) is made up of polynomials with a degree of 2 or less, meaning they can be expressed as \( ax^2 + bx + c \). These polynomials are treated similarly to coordinate vectors but in a more abstract form.
Understanding how to work with polynomials in a vector space involves grasping their structure and how they can be manipulated using operations like addition and scalar multiplication.
By mastering these concepts, one gains a powerful toolset to tackle more complex algebraic problems.
In our exercise, the vector space \( \mathscr{P}_{2} \) is made up of polynomials with a degree of 2 or less, meaning they can be expressed as \( ax^2 + bx + c \). These polynomials are treated similarly to coordinate vectors but in a more abstract form.
Understanding how to work with polynomials in a vector space involves grasping their structure and how they can be manipulated using operations like addition and scalar multiplication.
By mastering these concepts, one gains a powerful toolset to tackle more complex algebraic problems.